Timeline for Can we pin down the "recursively closed" levels of $L$ by a first-order theory?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2020 at 13:25 | comment | added | SSequence | would relate to any of this, since I don't have any understanding of it. That's why I don't know how this observation relates to the secondary question. But essentially $p$ would only be "slightly" larger than $\alpha$ though (e.g. $p$ would be very small compared $\alpha$-recursive well-orderings of $\alpha$). Note: Instead of posting another comment (with only one sentence), I deleted the last comment and copy-pasted with an added sentence towards the end. | |
| Aug 19, 2020 at 12:04 | comment | added | SSequence | Also, regarding your secondary question, I have a (likely) trivial question about terminology. Is "$L_\alpha$-definable well-ordering of $\alpha$" synonymous to "$\alpha$ (or $L_\alpha$?)-arithmetic well-orderings of $\alpha$"? Anyway, one observation is that there exist countable, recursively inaccessible and non-gandy ordinals $\alpha$ such that, even though $\alpha$ is non-gandy, there exists an ordinal $\alpha<p<\alpha^+$ such that the supremum of $p$-computable well-orderings of $\alpha$ would be exactly equal to $\alpha^+$. But I don't know how $L_\alpha$-definable [continued] | |
| Aug 19, 2020 at 11:21 | comment | added | SSequence | Also, given the definition, it seems you want all ordinals (and obviously many more too) of the form $\beta_l$ (with $l$ being a limit ordinal) to be categorized as "recursively closed". | |
| Aug 19, 2020 at 10:55 | comment | added | SSequence | I don't fully understand the question so I might be missing something. But writing $\beta_\alpha$ to denote the $\alpha$-th ordinal which is admissible (or a limit of those), it seems that by your definition $\beta_{\omega+1}$ would be "recursively closed". Same for $\beta_{\omega \cdot 2+1}$ etc. Is this intended (in the sense that it doesn't matter for the question)? Or perhaps you wanted to replace "iff for every admissible $\alpha<\gamma$" with "iff for every admissible (or limit of admissibles) $\alpha<\gamma$"? | |
| Aug 19, 2020 at 5:29 | history | asked | Noah Schweber | CC BY-SA 4.0 |